Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 244: 64

Answer

$(2a, 2\sqrt{a})$

Work Step by Step

$$f'(x) = \frac{{\sqrt {x - a} - \frac{x}{{2\sqrt {x - a} }}}}{{x - a}} = \frac{{\sqrt {x - a} - \frac{x}{{2\sqrt {x - a} }}}}{{x - a}} \cdot \frac{{2\sqrt {x - a} }}{{2\sqrt {x - a} }} = \frac{{2x - 2a - x}}{{2{{(x - a)}^{\frac{3}{2}}}}} = \frac{{x - 2a}}{{2{{(x - a)}^{\frac{3}{2}}}}}$$.This is zero when $x = 2a$, so there is a critical point at $(2a, 2\sqrt{a})$.
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